example of a Bezout domain that is not a PID

In the following example, ideals are considered to be of ??\mathbb{A} unless indicated otherwise via intersection with a subring of ??\mathbb{A}.

Let III be a finitely generated ideal of ??\mathbb{A}. Then there exists a positiveintegernnn and α1,…,αn∈?subscriptα1normal-…subscriptαn?\alpha_{1},\dots,\alpha_{n}\in\mathbb{A} with I=⟨α1,…,αn⟩Isubscriptα1normal-…subscriptαnI=\langle\alpha_{1},\dots,\alpha_{n}\rangle. Let K=ℚ⁢(α1,…,αn)Kℚsubscriptα1normal-…subscriptαnK=\mathbb{Q}(\alpha_{1},\dots,\alpha_{n}), and let ?Ksubscript?K\mathcal{O}_{K} denote the ring of integers of KKK. Then α1,…,αn∈?Ksubscriptα1normal-…subscriptαnsubscript?K\alpha_{1},\dots,\alpha_{n}\in\mathcal{O}_{K} and I∩?KIsubscript?KI\cap\mathcal{O}_{K} is an ideal of ?Ksubscript?K\mathcal{O}_{K}. Let hhh denote the class number of KKK. Then (I∩?K)h=⟨β⟩∩?KsuperscriptIsubscript?Khβsubscript?K(I\cap\mathcal{O}_{K})^{h}=\langle\beta\rangle\cap\mathcal{O}_{K} for some β∈?Kβsubscript?K\beta\in\mathcal{O}_{K}. Let L=K⁢(βh)LKhβL=K(\sqrt[h]{\beta}), and let ?Lsubscript?L\mathcal{O}_{L} denote the ring of integers of LLL. Then

Since unique factorization of ideals holds in ?Lsubscript?L\mathcal{O}_{L}, I∩?L=⟨βh⟩∩?LIsubscript?Lhβsubscript?LI\cap\mathcal{O}_{L}=\langle\sqrt[h]{\beta}\rangle\cap\mathcal{O}_{L}. Since ?K⊆?Lsubscript?Ksubscript?L\mathcal{O}_{K}\subseteq\mathcal{O}_{L} and α1,…,αn∈I∩?K⊆I∩?L=⟨βh⟩∩?Lsubscriptα1normal-…subscriptαnIsubscript?KIsubscript?Lhβsubscript?L\alpha_{1},\dots,\alpha_{n}\in I\cap\mathcal{O}_{K}\subseteq I\cap\mathcal{O}_%
{L}=\langle\sqrt[h]{\beta}\rangle\cap\mathcal{O}_{L}, there exist γ1,…,γn∈?Lsubscriptγ1normal-…subscriptγnsubscript?L\gamma_{1},\dots,\gamma_{n}\in\mathcal{O}_{L} with αj=γj⁢βhsubscriptαjsubscriptγjhβ\alpha_{j}=\gamma_{j}\sqrt[h]{\beta} for all positive integers jjj with j≤njnj\leq n. Thus, I=⟨α1,…,αn⟩=⟨γ1⁢βh,…,γn⁢βh⟩⊆⟨βh⟩Isubscriptα1normal-…subscriptαnsubscriptγ1hβnormal-…subscriptγnhβhβI=\langle\alpha_{1},\dots,\alpha_{n}\rangle=\langle\gamma_{1}\sqrt[h]{\beta},%
\dots,\gamma_{n}\sqrt[h]{\beta}\rangle\subseteq\langle\sqrt[h]{\beta}\rangle. Since I⊆⟨βh⟩IhβI\subseteq\langle\sqrt[h]{\beta}\rangle and I∩?L=⟨βh⟩∩?LIsubscript?Lhβsubscript?LI\cap\mathcal{O}_{L}=\langle\sqrt[h]{\beta}\rangle\cap\mathcal{O}_{L}, I=⟨βh⟩IhβI=\langle\sqrt[h]{\beta}\rangle. Hence, III is principal. It follows that ??\mathbb{A} is a Bezout domain.